20. (1) Solution: $\because e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{m}{m+2}} = \sqrt{\frac{2}{m+2}}$
Since $m > 1$, $\therefore 0 < e < \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$.
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$\therefore e \in \left(0, \frac{\sqrt{6}}{3}\right)$ & (2) Proof: Since the major axis length of the ellipse is $2\sqrt{m+2} = 4$, $\therefore m = 2$. \\
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The equation of line $BM$ is $y = -\frac{y_0}{4}(x-2)$, i.e., $y = -\frac{y_0}{4}x + \frac{1}{2}y_0$. & Substituting into the ellipse equation $x^2 + 2y^2 = 4$, \\
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By Vieta's formulas, $2x_1 = \frac{4(y_0^2 - 8)}{y_0^2 + 8}$, & 9 marks \\
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10 marks & \\
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$\therefore x_1 = \frac{2(y_0^2 - 8)}{y_0^2 + 8}$, $\therefore y_1 = \frac{8y_0}{y_0^2 + 8}$, $\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$ $\therefore \overrightarrow{OP} \cdot \overrightarrow{OM} = -2x_1 + y_0y_1 = -\frac{4(y_0^2 - 8)}{y_0^2 + 8} + \frac{8y_0^2}{y_0^2 + 8} = \frac{4y_0^2 + 32}{y_0^2 + 8} = 4 = \mathbf{2m}$. & 12 marks \\
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